8.2 Problems Involving Dry Friction Example 8.6 Determine the normal force P that must be exerted on the rack to begin pushing the 100kg pipe up the 20° incline. The coefficients of static friction at points of contact are (μs)A = 0.15 and (μs)B = 0.4.

8.2 Problems Involving Dry Friction View Free Body Diagram Solution The rack must exert a force P on the pipe due to force equilibrium in the x direction 4 unknowns 3 equilibrium equations and 1 frictional equation which apply at either A or B If slipping begins to occur at B, the pipe will roll up the incline If the slipping occurs at A, the pipe will begin to slide up the incline

8.2 Problems Involving Dry Friction Solution FBD of the rack

8.2 Problems Involving Dry Friction Solution Pipe rolls up incline

8.2 Problems Involving Dry Friction Solution Inequality does not apply and slipping occurs at A Pipe slides up incline Solving Check: no slipping occur at B

8.3 Wedges A simple machine used to transform an applied force into much larger forces, directed at approximately right angles to the applied force Used to give small displacements or adjustments to heavy load Consider the wedge used to lift a block of weight W by applying a force P to the wedge

8.3 Wedges FBD of the block and the wedge Exclude the weight of the wedge since it is small compared to weight of the block

8.3 Wedges Frictional forces F1 and F2 must oppose the motion of the wedge Frictional force F3 of the wall on the block must act downward as to oppose the block’s upward motion Location of the resultant forces are not important since neither the block or the wedge will tip Moment equilibrium equations not considered 7 unknowns - 6 normal and frictional force and force P

8.3 Wedges 2 force equilibrium equations (∑Fx = 0, ∑Fy = 0) applied to the wedge and block (4 equations in total) and the frictional equation (F = μN) applied at each surface of the contact (3 equations in total) If the block is lowered, the frictional forces will act in a sense opposite to that shown Applied force P will act to the right if the coefficient of friction is small or the wedge angle θ is large

8.3 Wedges Otherwise, P may have the reverse sense of direction in order to pull the wedge to remove it If P is not applied or P = 0, and friction forces hold the block in place, then the wedge is referred to as self-locking

8.3 Wedges Example 8.7 The uniform stone has a mass of 500kg and is held in place in the horizontal position using a wedge at B. if the coefficient of static friction μs = 0.3, at the surfaces of contact, determine the minimum force P needed to remove the wedge. Is the wedge self- locking? Assume that the stone does not slip at A.

8.3 Wedges View Free Body Diagram Solution Minimum force P requires F = μs NA at the surfaces of contact with the wedge FBD of the stone and the wedge On the wedge, friction force opposes the motion and on the stone at A, FA ≤ μsNA, slipping does not occur

8.3 Wedges Solution 5 unknowns, 3 equilibrium equations for the stone and 2 for the wedge

8.3 Wedges Solution Since P is positive, the wedge must be pulled out If P is zero, the wedge would remain in place (self-locking) and the frictional forces developed at B and C would satisfy FB < μsNB FC < μsNC